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Solution :

There are `4` sets possible with the numbers having difference multiple of `4`.<br>
`A = {1,5,9}`<br>
`B = {0,4,8}`<br>
`C = {2,6,10}`<br>
`D= {3,7}`<br>
If we do the sum of any two elements of set `A` and set `D`, it is not a multiple of `4`.<br>
If we do the sum of any two elements of set `B` and set `C`, it is a multiple of `4`.<br>
Now, we have to select `2` elements from each of the sets `B` and `C`.<br>
It can be done in `2**C(3,2)` ways.<br>
`:.` Required probability ` = (2C(3,2))/(C(11,2)) = (2**3)/(11**5) = 6/55`
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